Question: $f(x)=\cos(x+\pi)$ Find the first two nonzero terms in the Maclaurin series of $f$. Choose 1 answer: Choose 1 answer: (Choice A) A $-1+\frac{1}{2}x^2$ (Choice B) B $1+\frac{1}{2}x^2$ (Choice C) C $1-\frac{1}{2}x^2$ (Choice D) D $-x+\frac{1}{3!}x^3$ (Choice E) E $ x-\frac{1}{3!}x^3$
Solution: The first four terms of the Taylor series, centered at $~x=0\,$, for the function $~f(x)=\cos(x+\pi)~$ are given by the formula that appears below. $ f(0)+{f}\,^\prime(0)x+{f}\,^{\prime\prime}(0)\frac{x^2}{2!}+{f}\,^{\prime\prime\prime}(0)\frac{x^3}{3!}$ [Note that we use more terms than we might expect because the function $~f(x)=\cos(x+\pi)~$ and its derivatives can give us the value $~0~$ when evaluated at $~x=0\,$, and the problem asks for nonzero terms.] Below, we calculate the necessary derivatives and evaluate them at $~x=0\,$. $ f(x)=\cos(x+\pi) ~~~~~f(0)=-1$ $ f\,^\prime(x)=-\sin(x+\pi) \,f\,^\prime(0)=0$ $ f\,^{\prime\prime}(x)=-\cos(x+\pi) f\,^{\prime\prime}(0)=1$ $ f\,^{\prime\prime\prime}(x)=\sin(x+\pi) ~~~f\,^{\prime\prime\prime}(0)=0$ Hence, the desired terms are $ -1+0x+\frac{1}{2!}x^2$. This simplifies to $ -1+\frac{1}{2}x^2$.